Let $p\equiv 2\mod 3$ an odd prime number.
Prove that:
$p \mid (x^3+y^3) \implies p \mid (x+y)$ , for any integers $x,y$
$p\mid (x+y)(x^2-xy+y^2) \implies p\mid (x+y)$ or $p\mid (x^2-xy+y^2) $
if $p\mid (x+y)$ so problem solved
if $p\mid (x^2-xy+y^2)$ I can't find a contradiction.
We give two proofs, one by a Legendre symbol calculation, the second much simpler.
First Proof: If $p$ divides one of $x$ or $y$, then it must divide the other, and swe are finished. So suppose $p$ divides neither $x$ nor $y$. We show that $p$ cannot divide $x^2-xy+y^2$.
Suppose to the contrary that it does. Then $p$ divides $4x^2-4xy+4y^2$, that is, $p$ divides $(2x-y)^2+3y^2$.
Then $(2x-y)^2\equiv -3y^2\pmod{p}$. But $y$ is invertible modulo $p$. Multiplying through by the inverse of $y^2$, we find that the congruence $$w^2\equiv -3\pmod{p}$$ has a solution.
We reach a contradiction by showing that $-3$ is not a quadratic residue of $p$.
There are two cases, (i) $p\equiv 1\pmod{4}$ and (ii) $p\equiv -1\pmod{4}$.
For each case, we do a Legendre symbol calculation.
Case (i): We have $(-3/p)=(-1/p)(3/p)$. But $(-1/p)=1$, and by Quadratic Reciprocity $(3/p)=(p/3)=(2/3)=-1$. So $-3$ is not a quadratic residue of $p$.
The calculation for Case (ii) is very similar. Again, it turns out that $-3$ is not a quadratic residue of $p$.
Second proof: It is convenient to look at the equivalent problem of showing that if $p$ divides $x^3-y^3$, then $p$ divides $x-y$.
As in the first argument, we can assume that $y$ is not divisible by $p$. Then from the congruence $x^3\equiv y^3\pmod{p}$, multiplying through by $y^{-1}$, we get that $(xy^{-1})^3\equiv 1\pmod{p}$. We want to conclude that $x\equiv y\pmod{p}$.
Since $3$ and $\varphi(p)$ are relatively prime, the congruence $t^3\equiv 1\pmod{p}$ has precisely one solution, and we are finished.
If detail is needed, there exist integers $u$ and $v$ such that $3u+(p-1)v=1$. If $t^3\equiv 1\pmod{p}$, then we get the result from $t=t^1=t^{3u+(p-1)v}$.
Hint: Try to show that when $x^2+y^2-xy$ is divided by $3$, the remainder cannot be $2$. Also, $a^2\equiv0,1 \mod 3$
